Show that \(\maxspec(R) \subseteq \spec(R)\), and give an example where the containment is strict.
Solution
1. Every maximal ideal of \(R\) is prime.
Let \(\mathfrak m\) be a maximal ideal. Then \(R/\mathfrak m\) is a field, hence an integral domain. Therefore \(\mathfrak m\) is prime.
2. Hence \(\maxspec(R)\subseteq\spec(R)\).
By definition, \(\maxspec(R)\) is the set of maximal ideals and \(\spec(R)\) is the set of prime ideals. The inclusion follows from 1.
3. The inclusion can be strict.
Take \(R=\ZZ\). Since \(\ZZ\) is an integral domain, \((0)\) is a prime ideal. It is not maximal because \[ (0)\subsetneq (2)\subsetneq\ZZ. \] Thus \[ (0)\in\spec(\ZZ)\setminus\maxspec(\ZZ), \] so \(\maxspec(\ZZ)\subsetneq\spec(\ZZ)\).
4. Strictness is not automatic for every ring.
If \(R=k\) is a field, then \((0)\) is its only proper ideal, and it is both prime and maximal. Hence \[ \maxspec(k)=\spec(k)=\{(0)\}. \]